Not to be confused with "multiplication by a scalar" is dot product or scalar product , so named because the result is a scalar rather than a vector. This operations will have many uses in our work with vectors in a plane or in space. You see immediately that the operation exists in the same form for space as for the plane and that the useful theorems hold up in either case.
If we "say" the definition it is: The dot product of two vectors results in a unique scalar(real number) that is computed by summing the products of all corresponding entries ( that is, first with first and second with second, etc.) in the two vectors.
A more formal definition with some properties ( the author uses these to prove the first theorem ) is from the text:
See 11.3 Dot Product and Theorem 11.4 Properties of the Dot Product
We introduced the idea of norm of a vector before we had defined dot product but let us notice that norm is related to dot product as follows: the dot product of a vector with itself is the square of its norm. If we look back at the properties of dot product in our text we see it listed as #5. You should be able to verify this by simply computing the dot product of, let's use v=(a,b,c), with itself; observe that your result is the sum of the squares of a,b,and c which is in fact the square of the norm of v. The idea I'm expressing here is that dot product "induces" a norm on the space- an idea that is useful in general work on vector spaces.
An idea that is of more immediate use is given by a theorem which "says" : The cosine of the smaller undirected angle measure between two vectors is the number that results from the dot product of the two vectors divided by the product of the norms of those vectors. A more formal mathematical statement with a proof is from our text:
See Theorem 11.5 Angle Between Two Vectors; it follows Example 1
In the author's proof you can see the problem we could have with vector "sides" on the triangle. Because we have defined the difference of two vectors, u-v , and have the ability to compute norms of u,v, and u-v we are able to use the "law of cosines" because the length of the sides of the triangle are the norms of those vectors and consequently are real numbers.
As you go through the line in the proof, try to identify the property from Theorem 11.4 that is used to justify each "step" .
Next, your author defines orthogonal vectors as vectors with dot product equal zero. Then Theorem 10.5 justifies the statement that orthogonal vectors have an angle measure of 90 degrees between them. "Orthogonality" can be thought of as an extension of the idea of "perpendicular".
In the next example from the text you can observe that computing the cosine of the angle between two vectors is an exercise in computing dot product and vector norms.
Next I'll show two examples of the previous theorem at work: Measure of Direction Angles is one and the other is Vector Projection.
When doing mathematics for a line in a plane the tangent function was appropiate for computing the unique real number called the slope of the line. The tangent of the "angle of inclination" ( the smallest undirected angle between the line and the positive sense of the horizontal axis ) could be used to compute the slope of the line ( the ratio of "rise to run" ---"the opposite to the adjacent" ). The arctangent function can then be used to get the angle measure. The complement of that angle measure can be used for the angle between the vector and the positive sense of the vertical axis.
There will not be a unique real number for a line in space that allows us to compute the three angle measures between a line and the three coordinate axises. The idea of slope is for lines in a plane; for those in space we need a different approach.
Instead, the three angle measures can be computed using the cosine function, given that our previous theorem "employed" the cosine function. The angles to the coordinate axises are called direction anglesand the unique real number cosines of the angles will be called direction cosine numbers or just direction cosines.
Unit vectors i, j, and k on the coordinate axes and the Theorem 11.5 provide easy access to the direction cosine numbers whose arccosines produce the angle measures for the direction angles.
Finally we look at one more use of the theorem; it is the essential ingredient in the proof of the next theorem that provides a computation for producing a vector projection.
Before stating the theorem formally let's "talk it out" : the projection of one vector, say v , onto another vector w produces a vector; this vector is computed by using a unit vector in the direction of w and with the same sense as w ( that is, normalize vector w ) multiplied by a scalar. The scalar will be the dot product of v and w divided by the norm of w.
See 11.3 Projections and Vector Components
To understand the authors' definition of Vector Projection you will need to read the definition while looking at the diagram just below it; as they describe an item find it in the diagram.
When you read the proof that is given of Theorem 11.6, you may have trouble following the author's computation of k. You need to recognize that the author gets the norm of vector u multiplied by the cosine of theta by substituting for the dot product of u and v . He substitutes the product of the norms of u and v times the cosine of theta in place of the dot product of u and v. This is justified by the Theorem 11.5 ( cosine of the angle between vectors ).
The significance of the norm of u times the cosine of theta can be seen by looking at the diagram again and recognizing the result of right triangle trigonometry, namely, the length of the adjacent side to the angle theta is the product of the length of the hypotenuse and the cosine of theta; this is the length of the vector projection.
In high school mathematics and/or physics courses you may have decomposed a vector into horizontal and vertical components; this you could do using right triangle trigonometry. A significance of the work on projection is that the previous component work can now be stated in terms of vector projection and that the decompositions need not be done in the horizontal and vertical directions but rather in any direction designated by another vector and then in a direction orthogonal to it.
See Example 5 Decomposing a Vector into it's Components
Finally, this last example illustrates why we might want components other than vertical and horizontal.
Look at the following solved problems
in the Exercises at the end of
11.3 : 1,11,15,21,25,35,45,47,59,61
Then do the following unsolved
problems as homework: 2,12,16,22,26,46,48,62